The little orthogonality theorem consists of two relations that are reduced forms of the mathematical expression for the great orthogonality theorem.
1^{st} little orthogonality relation
The 1^{st} relation is derived from eq21 by letting and , resulting in
Since , we have and
Question
Show that .
Answer
Since the traces of a matrices of the same class are the same, eq22 is equivalent to
where

 is the number of classes.
 is the number of elements in the th class.
 , called the character of a class, is the trace of a matrix belonging to the th class of the th irreducible representation.
Eq23 is known as the 1^{st} little orthogonality relation.
Question
Determine whether the representations and of the point group are reducible or irreducible.
Answer
Since the 1^{st} little orthogonality relation is derived from the great orthogonality theorem, which pertains only to irreducible representations, characters of a representation that do not satisfy the relation belong to a reducible representation. Using eq23,
Therefore, is an irreducible representation of the point group, while is a reducible representation of the point group.
Let’s rewrite eq23 as
Eq24 has the form of the inner product of two vectors and in a dimensional vector space, with components and respectively. The components are functions of and the two vectors are orthogonal to each other when . A dimensional vector space is spanned by orthogonal vectors, each with components. Since the number of vectors and the number of components of each vector are denoted by and respectively, the number of irreducible representations of a group is equal to the number of classes of that group. We say that the irreducible representations of a group form a complete set of basis vectors in the dimensional vector space, with the components of each basis vector being .
2^{nd} little orthogonality relation
Consider the matrices and with entries and respectively.
Both are square matrices because the number of irreducible representations of a group is equal to the number of classes of that group. Comparing with eq24, the entries of the matrix product are given by:
Since and are square matrices and , then , i.e.
Eq24b is the 2^{nd} little orthogonality relation.
Question
Show that if and are square matrices and , then .
Answer
Using the determinant identities of and , we have , which implies that and are not zero and are therefore nonsingular. So,